Suppose 22.93mL of KMnO4 solution used in the question above (KMnO4 solution in above question 36.00mL and concentration of 0.02036M)are needed to oxidize Fe(2+) to Fe(3+) in a 0.4927g pill. What is the mass percent of FeSO4 (dot) 7H2O (molar mass 278.03g/mol) in the pill?
What I would do is take the concentration of KMnO4 and then using the 22.93mL find the mols and then divide the 0.4927g by them. This gives me the molar mass of the pill. Then i find the percent by dividing the molar mass of FeSO4 (dot) H2O by the pill molar mass to find the percent. Is that correct?
Thank you :)
Yes, your approach is correct! Here's a step-by-step explanation:
1. Start by calculating the number of moles of KMnO4 solution used. You have 22.93 mL of a 0.02036 M solution. To find the moles, use the formula: moles = volume (L) x concentration (M). Convert mL to L by dividing by 1000.
moles of KMnO4 = (22.93 mL / 1000 mL/L) x 0.02036 M
2. Calculate the moles of Fe(2+) oxidized by KMnO4. The balanced equation for the reaction is:
5 Fe(2+) + MnO4(-) + 8 H(+) -> 5 Fe(3+) + Mn(2+) + 4 H2O
From the balanced equation, you can see that every 1 mole of KMnO4 oxidizes 5 moles of Fe(2+). Therefore, the moles of Fe(2+) oxidized can be calculated as:
moles of Fe(2+) = 5 x moles of KMnO4
3. Determine the molar mass of the pill using the given mass. You have 0.4927 g of the pill, so divide this mass by the moles of Fe(2+) calculated in step 2 to get the molar mass of the pill.
molar mass of pill = 0.4927 g / moles of Fe(2+)
4. Finally, calculate the mass percentage of FeSO4 · 7H2O in the pill using the molar mass of FeSO4 · 7H2O (278.03 g/mol) and the molar mass of the pill calculated in step 3.
mass percentage of FeSO4 · 7H2O = (molar mass of FeSO4 · 7H2O / molar mass of pill) x 100%
By following these steps, you should be able to find the mass percent of FeSO4 · 7H2O in the given pill.